Question

The maximum and minimum distances of a satellite revolving in an elliptical orbit are in the ratio $3:1$. If the speed of the satellite at the nearest distance is $v$, then the speed at the farthest distance is

• A. $\frac{v}{6}$
• B. $3v$
• C. $\frac{v}{3}$
• D. $6v$
• E. $9v$

Hint:

Closer to center → higher speed (conservation of angular momentum).

Answer 1

The Correct Option is C

Concept:
• Angular momentum conservation: \[ mvr = \text{constant} \]

Step 1: Given ratio
\[ r_{max} : r_{min} = 3 : 1 \]

Step 2: Apply conservation
\[ v_{min} \cdot r_{max} = v_{max} \cdot r_{min} \] Let nearest speed $= v$, so: \[ v \cdot 1 = v_f \cdot 3 \]

Step 3: Solve
\[ v_f = \frac{v}{3} \] Final Conclusion:
Option (C)